Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 971.
Wednesday, January 29, 2014
Geometry Problem 971: Equilateral Triangle, Rectangle, Common Vertex, Sum of Right Triangles Areas
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http://imagizer.imageshack.us/v2/800x600q90/835/8nxu.png
ReplyDeleteDraw rectangular FDNM as per sketch
Note that area ( EBC) = area(ENC)
And Area(EAF)= area(FME)
This problem becomes problem 969.
Let z(P) be the complex number representing P.
ReplyDeleteLet z(C)=0, z(B)=−a, z(E)=−a+bi.
Then
z(F) = (−a+bi)(1/2 + √3/2 i) = (−1/2 a − √3/2 b) + (−√3/2 a + 1/2 b)i
z(A) = −a + (−√3/2 a + 1/2 b)i
z(D) = (−√3/2 a + 1/2 b)i
Thus
BE = b, BC = a
AF = −1/2 a + √3/2 b, AE = √3/2 a + 1/2 b
DC = √3/2 a − 1/2 b, DF = 1/2 a + √3/2 b
Area of ΔEBC = 1/2 ab
Area of ΔFAE = 1/8 (−√3 a² + 2 ab + √3 b²)
Area of ΔFDC = 1/8 (√3 a² + 2 ab − √3 b²)
Hence,
Area of ΔEBC = Area of ΔFAE + Area of ΔFDC
Sir, can you give me the idea how to get Z(F) Z(B) and Z(D)
Deletethx for a lot
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